Question

if one zero of the polynomial P(x) =(a square + 9) X square + 4 x + 68 is reciprocal of other find the value of a
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Answer 1

Step-by-step explanation:

$(a^{2}+9)x^{2} +4x+68=p(x)$

take the zeros as p and 1/p

therefore, $(a^{2} +9)p^{2}+4p+68=(a^{2} +9)1/p^{2} +4/p+68=0$

=> $(a^{2} +9)p^{2} +4p=(a^{2} +9)1/(p^{2} )+4/p=0$

=>$(a^{2}+9)p^{2}-(a^{2}+9)\frac{1}{p^{2}}+4p-\frac{4}{p}=0$

=> $[(a^{2}+9)(p^{2}-\frac{1}{p^{2}})]+\frac{4p^{2}-4}{p}=0$

=> $(a^{2}+9)(p^{2}-\frac{1}{p^{2}})=-\frac{4p^{2}-4}{p}$

the rest u can do it

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